Prove that $L\{\frac{2}{\sqrt \pi} \int_0^{\sqrt t}e^{-u^2}du\}=\frac{1}{s(\sqrt {s+1})}$ Question: 

Prove that $L\{\frac{2}{\sqrt \pi} \int_0^{\sqrt t}e^{-u^2}du\}=\frac{1}{s(\sqrt {s+1})}$

Note: $L$ stands for Laplace transform.  
My try:
$$L\left\{\frac{2}{\sqrt \pi} \int_0^{\sqrt t}e^{-u^2}du\right\}=\frac{2}{\sqrt \pi} L\left\{\int_0^{\sqrt t}e^{-u^2}du\right\} \tag{$*$} $$
Now, According to the rule $L\{\int_0^x f(u)du\}=\frac{1}{s}L\{f(x)\}$, I assumed $f(u)=e^{-u^2}$. From what it seems, $x$ is replaced with $\sqrt t$.  So, We have $L\{\int_0^{\sqrt t}e^{-u^2}du\}=\frac{1}{s} L\{f(\sqrt t)\}$ .  
Since $f(u)=e^{-u^2}$, We have $f(\sqrt t)=e^{-t}$.  
So, by all of these, $(*)$ is equal to $\frac{2}{\pi} \frac{1}{s} L\{e^{-t}\}=\frac{2}{\pi} \frac{1}{s}\frac{1}{s+1}$  which is not the thing that the question claims.  
What did I do wrong?
 A: Alternatively, one may just integrate by parts,
$$
\begin{align}
&L\left\{\frac{2}{\sqrt \pi} \int_0^{\sqrt t}e^{-u^2}du\right\}(s)
\\\\&=\frac{2}{\sqrt \pi}\int_{0}^{\infty}e^{-st}\left(\int_0^{\sqrt t}e^{-u^2}du\right)dt
\\\\&=\frac{2}{\sqrt \pi}\left[\frac{e^{-st}}{-s}\left(\int_0^{\sqrt t}e^{-u^2}du\right)\right]_{0}^{\infty}+\frac{2}{s\sqrt \pi}\int_{0}^{\infty}e^{-st}\left(\frac1{2\sqrt t}\cdot e^{-t}\right)dt
\\\\&=0+\frac{2}{s\sqrt \pi}\int_{0}^{\infty}e^{-(s+1)v^2}dv
\\\\&=\frac{1}{s(\sqrt {s+1})}.
\end{align}
$$
A: To apply the rule you quote, the variable $t$ should be the end point of the integration domain. Setting $u=\sqrt x$, the function to be transformed can be written as $$ \frac 1 {\sqrt\pi}\int_0^t e^{-x}\,\frac{dx}{\sqrt x} $$ Then the Laplace transform is
$$L(f)=\frac 1{s\sqrt\pi}L\left(\frac{e^{-x}}{\sqrt x}\right)$$ which can be calculated using the shift rule and the power law transform, giving the expected result.
A: Note that the function: $$f(t) = \frac{2}{\sqrt \pi} \int_{0}^{\sqrt t} e^{-u^2} \, du$$ is written as $\operatorname{erf}(\sqrt t)$. Now, we can use the series representation of the integrand to get: $$\begin{align} \operatorname{erf}(\sqrt t) = \frac{2}{\sqrt \pi} \int_{0}^{\sqrt t} \left[ 1- \frac{u^2}{1!}+ \frac{u^4}{4!} - \frac{u^6}{3!}+…\right]\, du\\ =\frac{2}{\sqrt \pi} \left[\frac{\sqrt t}{1}- \frac{t^{3/2}}{3\times 1!} + \frac{t^{5/2}}{5\times 2!} - \frac{t^{7/2}}{7\times 3!} + … \right]\end{align}$$ Thus, we have: $$\begin{align} L[\operatorname{erf}(\sqrt t)] = \frac{2}{\sqrt \pi} L\left[\frac{\sqrt t}{1}- \frac{t^{3/2}}{3\times 1!} + \frac{t^{5/2}}{5\times 2!} - \frac{t^{7/2}}{7\times 3!} + … \right] \\ = \frac{2}{\sqrt \pi} \left[ \frac{\Gamma(3/2)}{s^{3/2}} - \frac{\Gamma(5/2)}{s^{5/2}\times 3\times 1!} + \frac{\Gamma(7/2)}{s^{7/2}\times 5\times 2!} - …\right]\\ = \frac{1}{s^{3/2}}\left[1-\frac{1}{2s}+\frac{(-1/2)(-3/2)}{2!}\frac{1}{s^2} + \frac{(-1/2)(-3/2)(-5/2)}{3!}\frac{1}{s^3} + …\right] \\ = \frac{1}{s^{3/2}}\left(1+\frac{1}{s}\right)^{\frac{-1}{2}} = \frac{1}{s\sqrt{1+s}} \end{align}$$
A: $$
y = \frac 2 {\sqrt \pi} \int_0^w e^{-u^2} \, du, \quad w = \sqrt t.
$$
$$
\frac{dy}{dt} = \frac{dy}{dw} \cdot \frac{dw}{dt} = \frac 2 {\sqrt\pi} e^{-w^2} \cdot \frac 1 {2\sqrt t} = \frac 2 {\sqrt\pi} e^{-t} \cdot \frac 1 {2\sqrt t} = \frac{e^{-t}}{\sqrt{\pi t}}. 
$$
\begin{align}
\\[10pt]
f'(t) & = \frac{e^{-t}}{\sqrt{\pi t}} \tag 1 \\[10pt]
f(t) & = \frac 2 {\sqrt\pi} \int_0^{\sqrt t} e^{-u^2} \, du \tag 2 \\[10pt]
(\mathcal L (f'))(s) & = \underbrace{\int_0^\infty e^{-st} f'(t)\,dt = -f(0) + s\int_0^\infty e^{-st} f(t) \, dt}_\text{integration by parts} = -f(0) + s(\mathcal L f)(s). \tag 3
\end{align}
If you know the Laplace transform of line $(1)$ above, then you can use the conclusion of line $(3)$ to find the Laplace transform of line $(2).$
